Question

For Problems 1 to 12, if it is appropriate to do so, use the normal approximation to the binomial to calculate the indicated probabilities

n= 300,p= 0.95,P(280 <X< 290)

Answer 1

If X ~Bin (n,p )

then , normal approximation to binomial states that , if np>10 and n(1-p) >10 thenX approximately follows a normal distribution with mean = np and variance = np(1-p) .

Here np = 285 and n(1-p) = 15 . So condition satisfies .

So, we can say approximately X~N( 285 , 3.775²)

So,

P[280 < X < 290] PE 280 - 285 3.775 X – 285 3.775 290 – 285 3.775 = P[-1.3245 <2<1.3245] = 20(1.3245) - 1 = 0.8147